Write an equation in the form $y=a(x-p)^{2}+q$ for a parabola congruent to $y=-5 x^{2}$ with a minimum of -6 and an axis of symmetry of $x=2$. $a=-5$

Step 1 ：Given the equation of the parabola in the form \(y = ax^2\), we want to write it in the form \(y = a(x-p)^2 + q\).

Step 2 ：Since the parabola is congruent to the given one, the value of \(a\) is the same in both equations, which is -5.

Step 3 ：The minimum of the parabola is -6, which is the value of \(q\).

Step 4 ：The axis of symmetry of the parabola is \(x = 2\), which is the value of \(p\).

Step 5 ：Substitute these values into the equation \(y = a(x-p)^2 + q\) to get the equation of the parabola.

Step 6 ：So, the equation of the parabola is \(y = -5(x - 2)^2 - 6\).

Step 7 ：\(\boxed{y = -5(x - 2)^2 - 6}\) is the final answer.